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Bond [772]
3 years ago
11

Why does the Sun appear so big, bright, and hot if it is only an average sized star?

Physics
2 answers:
Oliga [24]3 years ago
4 0

Answer:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller. The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Elena L [17]3 years ago
3 0

Answer:

The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Explanation:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller.

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kodGreya [7K]

To solve this problem we will apply the given concept for torque which explains the relationship between the force applied and the distance to a given point. Mathematically this relationship is given as

\tau = F*d \rightarrow F = \frac{\tau}{d}

Where,

\tau = Torque

F = Force

d = Distance

Our values are given as,

\tau = 7.0Nm , d = 0.55m

Therefore replacing we have that the force is

F = \frac{7}{0.55}

F = 12.72N

Therefore the least amount of force that you must exert is 12.72N

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A car slows down from 65 km/s to 30 km/s in 5 seconds. What is its acceleration?
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A large aquarium of height 5 m is filled with fresh water to a depth of D = 1.80 m. One wall of the aquarium consists of thick p
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To solve the problem we will first start considering the Pressure given the hydrostatic definition of the product between the density, the gravity and the depth. We will define the area where the liquid acts and later we will use the definition of the force as a product between the pressure and the area to calculate the force given in the two depths. The gauge pressure at the depth x will be

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This pressure acts on the strip of area

dA = 7.6dx

The force acting on that strip is given by,

dF = (P-P_a)dA

dF = \rho g xdA

dF = 7.6 \rho g x dx

To evaluate the force, we will then consider the integral of the pressure as a function of the Area, or the integral of the previously found terms.

F = \int_0^x 7.6 \rho g x dx

F = 3.8 \rho g x^2

Evaluating at the initial depth of 1.8m and the final depth of 4.4 we have then that,

F_{1.8} = 3.8 (998)(9.8)(1.8)^2 = 120416.28N

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F = F_{4.4}-F_{1.8}

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